Leetcode:464. Can I Win

Description

In the “100 game,” two players take turns adding, to a running total, any integer from 1..10. The player who first causes the running total to reach or exceed 100 wins.

What if we change the game so that players cannot re-use integers?

For example, two players might take turns drawing from a common pool of numbers of 1..15 without replacement until they reach a total >= 100.

Given an integer maxChoosableInteger and another integer desiredTotal, determine if the first player to move can force a win, assuming both players play optimally.

You can always assume that maxChoosableInteger will not be larger than 20 and desiredTotal will not be larger than 300.

Example
Input:
maxChoosableInteger = 10
desiredTotal = 11
Output:
false
Explanation:
No matter which integer the first player choose, the first player will lose.
The first player can choose an integer from 1 up to 10.
If the first player choose 1, the second player can only choose integers from 2 up to 10.
The second player will win by choosing 10 and get a total = 11, which is >= desiredTotal.
Same with other integers chosen by the first player, the second player will always win.

解题思路

题意是1-M的数中各轮流抽一个数，两个人抽的数的总和为T或者谁的数加上超过100的为winner。
令M = maxChoosableInteger , T = desiredTotal.
0 : not calculated yet;
1 : current player can win;
-1: current player can’t win.
There are several checks to be done at initial state k = 0 for early termination so we won’t waste our time for DFS process:

1. if T < 2, obviously, the first player wins by simply picking 1.
2. if the sum of entire pool S = M*(M+1)/2 is less than T, of course, nobody can reach T.
3. if the sum S == T, the order to pick numbers from the pool is irrelevant. Whoever picks the last will reach T. So the first player can win iff M is odd.（M为奇数时才能赢）

class Solution {
public:

    bool canIWin(int M, int T) {
      int S = M*(M+1)/2; // sum of entire pool
      return T<2? true : S<T? false : S==T? M%2 : dfs(M,T,0);
    }

    bool dfs(int M, int T, int k) {
      if (T<=0 || m[k]) return T>0 && m[k]>0; // memorization or total reached by opponent
      for (int i = 0; i < M; ++i)
        if (!(k&1<<i) && !dfs(M, T-i-1, k|1<<i)) return m[k] = 1; // current player wins
      return !(m[k] = -1); // current player can't win
    }

    int m[1<<20] = {}; // m[key]: memorized result when pool state = key
};

或者

class Solution {
public:
　　bool canIWin(int maxChoosableInteger, int desiredTotal) {
　　　　if (maxChoosableInteger >= desiredTotal) return true;
　　　　int sum = ((maxChoosableInteger + 1) * maxChoosableInteger) >> 1;
　　　　if (sum < desiredTotal) return false;
　　　　mp = vector<int>(1 << maxChoosableInteger, -1);
　　　　return canWin(0, maxChoosableInteger, desiredTotal);
　　}
private:
　　vector<int> mp;
　　bool canWin(int used, const int &maxChoosableInteger, int desiredTotal) {
　　　　if (mp[used] != -1) return mp[used];
　　　　for (int i = maxChoosableInteger, bits = 1 << (maxChoosableInteger - 1); i >= 1; --i, bits >>= 1) {
　　　　　　if ((used & bits) != 0) continue;
　　　　　　if (i >= desiredTotal || !canWin(used | bits, maxChoosableInteger, desiredTotal - i)) {
　　　　　　　　mp[used] = 1;
　　　　　　　　return true;
　　　　　　}
　　　　}
　　　　mp[used] = 0;
　　　　return false;
　　}
};